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Machine M and Machine N working alone at their constant rate [#permalink]
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26 Jul 2014, 10:07
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Machine M and Machine N working alone at their constant rates, non stop, produced 6000 and 8000 nails respectively. Did machine M work longer than machine N? (1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate (2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate
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Re: Machine M and Machine N working alone at their constant rate [#permalink]
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26 Jul 2014, 10:37
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Machine M and Machine N working alone at their constant rates, non stop, produced 6000 and 8000 nails respectively. Did machine M work longer than machine N?(1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate. If the rate of M is 2,000 nails per hour and the rate of N is 4,000 nails per hour, then to produce 6,000 nails, M needs 3 hours and to produce 8,000 nails N, needs 2 hours. In this case M worked longer than N. If the rate of M is 6,000 nails per hour and the rate of N is 8,000 nails per hour, then to produce 6,000 nails, M needs 1 hours and to produce 8,000 nails N, needs 1 hour. In this case M did not work longer than N. Not sufficient. (2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate. In the time M needs to produce 6,000 nails, N can produce 12,000 nails, thus it can produce 8,000 nails in less time than M can produce 6,000 nails. Sufficient. Answer: B.
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Re: Machine M and Machine N working alone at their constant rate [#permalink]
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05 Sep 2014, 07:22
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Machine M and Machine N working alone at their constant rates, non stop, produced 6000 and 8000 nails respectively. Did machine M work longer than machine N? Let N is number of nails machine N produced. Let M is number of nails machine M produced. And x and y are respective time each machine took to produce 6000 and 8000 nails. (1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate Given N= M + 2000 Mx = 6000 (1) Ny = 8000 ==> (M+2000)y=8000 (2) Solve 1 and 2 to find ratio of x/y. \((\frac{6000}{x}+2000)y=8000\) We can not reduce this equation in \(x/y\) form. Thus, (1) insufficient. (2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate. N=2M Mx = 6000 (1) Ny = 8000 ==> 2My=8000 (2) \(2y\frac{6000}{x}=8000\) Solve for \(x/y\); we can calculate x:y. Thus, statement (2) is sufficient.
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Re: Machine M and Machine N working alone at their constant rate [#permalink]
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07 Sep 2014, 09:33
Hi Bunuel,
Why are we rounding off the number of hours here? Am i missing something in the question, why arent we considering minutes, for which im getting the first statement as being sufficient. Thanks



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Re: Machine M and Machine N working alone at their constant rate [#permalink]
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07 Sep 2014, 09:43



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Re: Machine M and Machine N working alone at their constant rate [#permalink]
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07 Sep 2015, 04:15
Bunuel wrote: Machine M and Machine N working alone at their constant rates, non stop, produced 6000 and 8000 nails respectively. Did machine M work longer than machine N?
(1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate.
If the rate of M is 2,000 nails per hour and the rate of N is 4,000 nails per hour, then to produce 6,000 nails, M needs 3 hours and to produce 8,000 nails N, needs 2 hours. In this case M worked longer than N.
If the rate of M is 6,000 nails per hour and the rate of N is 8,000 nails per hour, then to produce 6,000 nails, M needs 1 hours and to produce 8,000 nails N, needs 1 hour. In this case M did not work longer than N.
Not sufficient.
(2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate. In the time M needs to produce 6,000 nails, N can produce 12,000 nails, thus it can produce 8,000 nails in less time than M can produce 6,000 nails. Sufficient.
Answer: B. The highlighted part: If the question is to produce 6000 nails then definitely M needs 1 hour but N with a rate of 8000 nails per hour would need less than an hour right? Please clarify if possible. Thank you.



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Re: Machine M and Machine N working alone at their constant rate [#permalink]
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07 Sep 2015, 05:06
earnit wrote: Bunuel wrote: Machine M and Machine N working alone at their constant rates, non stop, produced 6000 and 8000 nails respectively. Did machine M work longer than machine N?
(1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate.
If the rate of M is 2,000 nails per hour and the rate of N is 4,000 nails per hour, then to produce 6,000 nails, M needs 3 hours and to produce 8,000 nails N, needs 2 hours. In this case M worked longer than N.
If the rate of M is 6,000 nails per hour and the rate of N is 8,000 nails per hour, then to produce 6,000 nails, M needs 1 hours and to produce 8,000 nails N, needs 1 hour. In this case M did not work longer than N.
Not sufficient.
(2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate. In the time M needs to produce 6,000 nails, N can produce 12,000 nails, thus it can produce 8,000 nails in less time than M can produce 6,000 nails. Sufficient.
Answer: B. The highlighted part: If the question is to produce 6000 nails then definitely M needs 1 hour but N with a rate of 8000 nails per hour would need less than an hour right? Please clarify if possible. Thank you. How does N, at the rate of 8,000 nails per hour, need more than an hour to produce 8000 nails? It will need exactly 1 hour. Again, the question asks: did machine M (which produced 6000 nails) work longer than machine N (which produced 8000 nails)? If the rate of M is 6,000 nails per hour and the rate of N is 8,000 nails per hour, then to produce 6,000 nails, M needs 1 hours and to produce 8,000 nails N, needs 1 hour. In this case M did not work longer than N.
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Re: Machine M and Machine N working alone at their constant rate [#permalink]
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07 Sep 2015, 08:06
Bunuel wrote: earnit wrote: Bunuel wrote: Machine M and Machine N working alone at their constant rates, non stop, produced 6000 and 8000 nails respectively. Did machine M work longer than machine N?
(1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate.
If the rate of M is 2,000 nails per hour and the rate of N is 4,000 nails per hour, then to produce 6,000 nails, M needs 3 hours and to produce 8,000 nails N, needs 2 hours. In this case M worked longer than N.
If the rate of M is 6,000 nails per hour and the rate of N is 8,000 nails per hour, then to produce 6,000 nails, M needs 1 hours and to produce 8,000 nails N, needs 1 hour. In this case M did not work longer than N.
Not sufficient.
(2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate. In the time M needs to produce 6,000 nails, N can produce 12,000 nails, thus it can produce 8,000 nails in less time than M can produce 6,000 nails. Sufficient.
Answer: B. The highlighted part: If the question is to produce 6000 nails then definitely M needs 1 hour but N with a rate of 8000 nails per hour would need less than an hour right? Please clarify if possible. Thank you. How does N, at the rate of 8,000 nails per hour, need more than an hour to produce 8000 nails? It will need exactly 1 hour. Again, the question asks: did machine M (which produced 6000 nails) work longer than machine N ( which produced 8000 nails)? If the rate of M is 6,000 nails per hour and the rate of N is 8,000 nails per hour, then to produce 6,000 nails, M needs 1 hours and to produce 8,000 nails N, needs 1 hour. In this case M did not work longer than N. Got it. Its not the common number that the machines are expected to produce. That part slipped out when exploring the options. Key take away: if there is a common number required to produce then even (1) would be sufficient. But not in the above case.



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Machine M and Machine N working alone at their constant rate [#permalink]
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08 Sep 2015, 07:38
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution. Machine M and Machine N working alone at their constant rates, non stop, produced 6000 and 8000 nails respectively. Did machine M work longer than machine N? (1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate (2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate In the original condition, r1*t1=6000, r2*t2=8000 and throw some transformation we have t1=6000/r1, t2=8000/r2 and 6000/r1>8000/r2, 3r2>4r1?, therefore the question is all about comparing r1 and r2. We have 4 variables (r1,r2,t1,t2), 2 equations (r1*t1=6000, r2*t2=8000) therefore we need 2 more equations and therefore C is likely the answer. Using both 1) & 2) together, r2=r1+2000, r2=2r1 gives us r1=2000, r2=4000 and thus C is the answer. But such trivial conditions are rarely the answer, so we might try it separately again. Using 1), 2)separately, (from Common mistake type 4(A), in case of 2) substituting r2=2r1 to 3r2>4r1? gives us 3*2r1>4r1?, 6>4. Therefore the answer is yes, and the condition is sufficient. Therefore the answer is B. (when looking at conditions 1), 2) if one is given by value and the other is given by ratio, the one with ratio is usually the answer. That's why we just calculated using con 1) here. )
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Re: Machine M and Machine N working alone at their constant rate [#permalink]
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21 Nov 2016, 16:37
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Re: Machine M and Machine N working alone at their constant rate [#permalink]
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31 Dec 2016, 03:52
Here is my approach
Is 6000/(rate of m = rm) > 8000 / (rate of n = rn) ? <> Is 6/rm > 8/rn? <> Is 6 rn > 8 rm ?
St. 1: rn = rm + 2000 > Is 6 rm + 12000 > 8 rm ? Here we get a condition, i.e. rm < 6000 that is not verified a priori > NOT SUFF
St. 2: rn = 2 rm > Is 12 rm > 8 rm ? Yes because rm >0 > SUFF
B




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